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An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
A
$\frac{1}{v^4}$
B
$\frac{1}{Ze}$
C
$v^2$
D
$\frac{1}{m}$
Detailed Solution
At the distance of closest approach $r_0$, all the kinetic energy of the alpha particle is converted into electrostatic potential energy.
$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r_0}$
$r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{\frac{1}{2}mv^2} = \frac{4Ze^2}{4\pi\varepsilon_0mv^2}$
So $r_0 \propto Ze$, $r_0 \propto \frac{1}{v^2}$ and $r_0 \propto \frac{1}{m}$.
Among the given choices, the correct proportionality is $r_0 \propto \frac{1}{m}$.
$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r_0}$
$r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{\frac{1}{2}mv^2} = \frac{4Ze^2}{4\pi\varepsilon_0mv^2}$
So $r_0 \propto Ze$, $r_0 \propto \frac{1}{v^2}$ and $r_0 \propto \frac{1}{m}$.
Among the given choices, the correct proportionality is $r_0 \propto \frac{1}{m}$.